Minimum distance between BST nodes

Time: O(N); Space: O(H); easy

Given a Binary Search Tree (BST) with the root node root,

return the minimum difference between the values of any two different nodes in the tree.

Example 1:

Input: root = {TreeNode} [4,2,6,1,3,null,null]

Output: 1

Explanation:

  • Note that root is a TreeNode object, not an array.

  • The given tree [4,2,6,1,3,null,null] is represented by the following diagram:

        4
   / \
  2   6
 / \
1   3

  • While the minimum difference in this tree is 1, it occurs between node 1 and node 2, also between node 3 and node 2.

Notes:

  • The size of the BST will be between 2 and 100.

  • The BST is always valid, each node’s value is an integer, and each node’s value is different.

[1]:

class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

[2]:

class Solution1(object):
    """
    Time: O(N)
    Space: O(H)
    """
    def minDiffInBST(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        def dfs(node):
            if not node:
                return
            dfs(node.left)
            self.result = min(self.result, node.val-self.prev)
            self.prev = node.val
            dfs(node.right)

        self.prev = float('-inf')
        self.result = float('inf')
        dfs(root)

        return self.result

[3]:

s = Solution1()

root = TreeNode(4)
root.left = TreeNode(2)
root.right = TreeNode(6)
root.left.left = TreeNode(1)
root.left.right = TreeNode(3)
assert s.minDiffInBST(root) == 1